Why $\vec v \times \vec p = 0$ for rotating particle ? 

A particle of mass $m$ moves in the $XY$ plane with a velocity $v$ along the straight line $AB.$ If the angular momentum of the particle with respect to origin $O$ is $L_A$ when it is at $A$ and $L_B$ when it is at $B,$ then

Define angular momentum.

A particle of mass $2\, kg$ is moving such that at time $t$, its position, in meter, is given by $\overrightarrow r \left( t \right) = 5\hat i - 2{t^2}\hat j$ . The angular momentum of the particle at $t\, = 2\, s$ about the origin in $kg\, m^{-2}\, s^{-1}$ is

A particle is moving along a straight line parallel to $x$-axis with constant velocity. Find angular momentum about the origin in vector form

The time dependence of the position of a particle of mass $m = 2$ is given by $\vec r\,(t)\, = \,2t\,\hat i\, - 3{t^2}\hat j$ Its angular momentum with respect to the origin at time $t = 2$ is